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At 0^(@)C ice and water are in equilibri...

At `0^(@)C` ice and water are in equilibrium and `DeltaH=6kJ" "mol^(-1)` for this process:
`H_(2)OhArrH_(2)O(l)`
The values of `DeltaS and DeltaG` for conversion of ice into liquid water at `0^(@)C` are:

Text Solution

Verified by Experts

Since the given process inequilibrium , `Delta G = 0`
Putting this value in the relationship , `Delta G = DeltaH - T DeltaS ` weget
`0= Delta H-T Delta S` or `T DeltaS = DeltaH ` or `Delta S = (DeltaH)/(T)`
We are given `DeltaH = 6.0 kJ mol^(-1) = 6000 J mol^(-1) `and `T = 0^(@) C = 273 K`
`:. Delta S = ( 6000 J mol^(-1))/( 273K ) = 21.98 J K^(-1) mol^(-1)`
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